Bhagya

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.2

Polynomial in ascending order calculator.

Question 1.
The scores in Mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Solution:
Arranging the given scores in ascending order, we have 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode of given data is that value of observation which occurs for the most number of times i.e., 20. 5.
Median of the given data is the middle observation when the data is arranged in ascending order i.e., 8^th term = 20
Hence, mode of data = median of data.

Question 2.
The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Find the mean, mode and median of this data. Are the three same?
Solution:
Arranging the given scores in an ascending order, we have

As 15 occurs for the most number of times.
∴ Mode = 15
Median = Middle term = 15
∴ All three mean, mode and median are not same.

Question 3.
The weights (in kg.) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Solution:
Arranging the given weights in ascending order, we have 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45,47, 50
(i) Mode = 38 and 43
Median = Middle term = 40
(ii) Yes, there are 2 modes for the given data i.e., 38 and 43.

Question 4.
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:
Arranging the given data in an ascending order, we have
12, 12, 13, 13, 14, 14, 14, 16, 19
14 occurs for the most number of times
∴ Mode = 14
Median = Middle observation = 14

Question 5.
Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6,4,3,8,9,12,13,9 has mean 9.
Solution:
(i) True, as mode of a given data is that value of observation which occurs for the most number of times. Therefore, it is one of the observations given in the data.
(ii) False, because mean may or may not be one of the numbers in the data.
(iii) False, because median may or may not be one of the numbers in the data
(iv) False, because

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Solution:

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm and breadth = 3 cm
Area = Length × Breadth = (5.7 × 3) cm² = 17.1 cm²

Question 3.
Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
We know that when a decimal number is multiplied by 10, 100, 1000, the decimal point in the product is shifted to the right by as many places as there are zeros.
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

For most, thinking of 1/2 as a decimal will make it easier to determine the square root.

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered in 1 litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = (10 × 55.3) km = 553 km

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Solution:

MP Board Class 7th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5

What is 5/8 as a decimal? Step-by-step solution to 5/8 as a decimal.

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:

which is the required solution.

Question 2.
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)
Solution:

⇒ n = 6, which is the required solution.

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:

⇒ x = -5, which is the required solution.

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:

⇒ x = 8, which is the required solution.

Question 5.
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:

⇒ t = 2, which is the required solution.

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5(2t + 1)
Solution:
We have, 3(t – 3) = 5(2t + 1)
⇒ 31 – 9 = 101 + 5
⇒ – 9 – 5 = 101 – 31 ⇒ -14 = 71
⇒ 1 = – 2, which is the required solution.

Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
We have,
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
⇒ 15 y – 60 – 2y + 18 + 5y + 30 = 0
⇒ 18y – 12 = 0 ⇒ 18y = 12

⇒ y = \(\frac{2}{3}\), which is the required solution.

Question 9.
3(5z – 7) – 2(9z -11) = 4(8z – 13) – 17
Solution:
We have,
3(5z – 7) – 2(9z -11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
⇒ 15z – 18z – 32z = 21 – 22 – 52 – 17 – 70
⇒ -35z = -70 ⇒ z = \(\frac{-70}{-35}\)
⇒ z = 2, which is the required solution.

Question 10.
0.25 (4f- 3) = 0.05(10f – 9)
Solution:
We have, 0.25(4f – 3) = 0.05(10f – 9)
⇒ f – 0.75 = 0.5f – 0.45

⇒ f = 0.6, which is the required solution.

MP Board Class 8th Maths Solutions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4

प्रश्न 1.
निर्धारित कीजिए कि क्या निम्नलिखित प्रकार से परिभाषित प्रत्येक संक्रिया से एक द्विआधारी संक्रिया प्राप्त होती है या नहीं। उस दशा में जब एक द्विआधारी संक्रिया नहीं है, औचित्य भी बतलाइए।
(i) Z⁺ में, a * b = a – b द्वारा परिभाषित संक्रिया
(ii) Z⁺ में, a* b = ab द्वारा परिभाषित संक्रिया
(iii) R में, संक्रिया *, a* b = ab² द्वारा परिभाषित
(iv) Z⁺ में, संक्रिया *, a* b = |a – b| द्वारा परिभाषित
(v) Z⁺ में, संक्रिया *, a* b = a द्वारा परिभाषित
हल:
(i) Z⁺ में, a* b = a – b द्वारा परिभाषित संक्रिया है
यदि a > b, a * b = a – b ϵ Z⁺
परन्तु यदि a < b, a * b = a – b < 0, Z⁺ में नहीं है।
अत:* संक्रिया द्विआधारी संक्रिया नहीं है।

(ii) Z⁺ पर * संक्रिया, a * b = ab द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ ⇒ a और b दोनों धनात्मक हैं।
a * b = ab भी धनात्मक है।
ab ϵ Z⁺
अतः यह संक्रिया द्विआधारी है।

(iii) R पर * संक्रिया a* b = ab⁺ द्वारा परिभाषित है।
यदि a, b ϵ R, ab² भी R* में है।
अतः यह संक्रिया द्विआधारी है।

(iv) Z⁺ पर * संक्रिया a * b = |a – b| द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ , |a – b | ϵ Z⁺
अंत: यह संक्रिया द्विआधारी है।

(v) Z⁺ पर * संक्रिया a* b = a द्वारा परिभाषित है।
यदि a, b ϵ Z⁺, ∴ a * b = a ϵ Z⁺
अत: यह संक्रिया द्विआधारी है।

प्रश्न 2.
निम्नलिखित परिभाषित प्रत्येक द्विआधारी संक्रिया के लिए निर्धारित कीजिए कि क्या द्वि आधारी क्रमविनिमेय है तथा क्या साहचर्य है।
(i) Z में, a * b = a – b द्वारा परिभाषित
(ii) Q में, a * b = ab + 1 द्वारा परिभाषित
(iii) Q में, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित
(iv) Z⁺ में, a * b = 2^ab द्वारा परिभाषित
(v) Z⁺ में, a* b = a^b द्वारा परिभाषित
(vi) R – { – 1} में, a* b = \(\frac{a}{b+1}\) द्वारा परिभाषित
हल:
(i) Z पर संक्रिया a* b = a – b द्वारा परिभाषित है।
(a) यदि a * b = a – b और b * a = b – a
परन्तु a – b ≠ b – a ⇒ a* b + b * a
∴ यह संक्रिया क्रमविनिमेय नहीं है।

(b) यदि a * (b * c) = a * (b – c) = a * (b – c)
a – (b – c) = a – b + c
(a * b) * c = (a – b) * c = a – b – c
स्पष्ट है कि a * (b * c) (a * b) * c
∴ संक्रिया साहचर्य नहीं है। अतः संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य है।

(ii) Q पर * संक्रिया, a * b = ab + 1 से परिभाषित है।
(a) a * b = ab + 1, b * a = ba + 1 = ab + 1
∴ a * b = b * a
∴ यह संक्रिया क्रमविनिमेय द्विआधारी है।

(b) यदि a * (b * c) = a * (bc + 1) = a (bc + 1) + 1
= abc + a + 1
(a * b) * c= (ab + 1) * c =(ab + 1)c + 1
= abc + c + 1
∴ (a * b) * c ≠ a + (b + c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(iii) Q पर * संक्रिया, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया है।
अतः यह संक्रिया क्रमविनिमेय और साहचर्य दोनों हैं।

(iv) Z⁺ पर * संक्रिया a* b = 2^ab से परिभाषित है।
(a) ∴ a * b = 2^ab, b * a = 2^ba = 2^ab
⇒ a * b = b * a
अतः संक्रिया क्रमविनिमेय संक्रिया है।

(b) a * (b * c) = a * 2^bc = a^a.2^bc
(a * b) * c = 2^ab * c = 2^2ab.c
∴ a * (b * c) ≠ * (a * b) *c
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(v) Z⁺ पर * संक्रिया, a * b = a^b से परिभाषित है।
(a) a * b = a^b, b * a = b^a
∴ a * b + b* a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c)=a * b^c = a^(bc)
(a * b) * c = ad * c = a^(b)c = a^bc
∴ (a * b) * c * a * (b * c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य

(vi) R – {-1} पर * संक्रिया, a * b = \(\frac{a}{b+1}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया क्रमविनिमेय है और न ही साहचर्य है।

प्रश्न 3.
समुच्चय {1, 2, 3, 4,5} में a ^ b= निम्नतम {a, b} द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए। संक्रिया के लिए संक्रिया सारणी लिखिए।
हल:
समुच्चय {1, 2, 3, 4, 5} पर संक्रिया ^ सारणी निम्न है-

प्रश्न 4.
समुच्चय {1, 2, 3, 4, 5} में, निम्नलिखित संक्रिया सारणी (सारणी 1.2) द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए तथा
(i) (2 * 3) * 4 तथा 2 * (3 * 4) का परिकलन कीजिए।
(ii) क्या * क्रम विनिमेय है?
(iii) (2 * 3) * (4 * 5) का परिकलन कीजिए। (संकेत : निम्न सारणी का प्रयोग कीजिए।)

हल:
(i) दी गई सारणी से
(2 * 3) * 4 = 1 * 4 = 1
तथा 2 * (3 * 4) = 2 * 1 = 1

(ii) माना a, b ϵ {1, 2, 3, 4, 5}
∴ सारणी से, a * a = a (a ≠ b) तथा a, b विषम संख्या है
a * b = b * a = 1
2 * 4 = 4 * 2 = 2 जहाँ a तथा b सम संख्या तथा a ≠ b
अतः a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) सारणी से,
(2 * 3) * (4 * 5) = 1 *1
= 1

प्रश्न 5.
मान लीजिए कि समुच्चय {1, 2, 3, 4, 5} में एक द्विआधारी संक्रिया *’, a *’ b = a तथा b का HCF द्वारा परिभाषित है। क्या संक्रिया *’ उपर्युक्त प्रश्न 4 में परिभाषित संक्रिया * के समान है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
यहाँ समुच्चय {1, 2, 3, 4, 5} संक्रिया a*’ b H.C.F. a, b द्वारा परिभाषित है।
इस संक्रिया की निम्न सारणी दी गयी है-

प्रश्न 4 में दी गई सारणी और यह सारणी समान है।
अतः संक्रिया *’ तथा * समान है।

प्रश्न 6.
मान लीजिए कि N में एक द्विआधारी संक्रिया, a * b = a तथा b का LCM द्वारा परिभाषित है। निम्नलिखित ज्ञात कीजिए:
(i) 5 * 7, 20 * 16
(ii) क्या संक्रिय * क्रम विनिमेय है?
(iii) क्या * साहचार्य है?
(iv) N में * का तत्समक अवयव ज्ञात कीजिए।
(v) N के कौन-से अवयव * संक्रिया के लिए व्युत्क्रमणीय है?
हल:
द्विआधारी संक्रिया (Binary Operations) * इस प्रकार परिभाषित है कि
a * b = a तथा b का L.C.M.
(i) 5 * 7 =5 तथा 7 का L.C.M.
= 35
तथा 20 * 16 = 20 तथा 16 का L.C.M.
= 80

(ii) a * b = a तथा b का L.C.M.
= b तथा a का L.C.M
a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) a * (b * c) = a * (b तथा c का L.C.M.)
= a तथा (b तथा c का L.C.M.) का L.C.M.
= a, b तथा c का L.C.M.
इसी प्रकार
(a * b) * c =(a तथा b का L.C.M.) * c
=a, b, c of L.C.M.
⇒ a *(b * c) =(a * b) * c
अतः द्विआधारी संक्रिया * साहचर्य है।

(iv) N में * संक्रिया की तत्समक अवयव 1 है।
∵ 1 * a = a * 1 = a
= 1 तथा a का L.C.M.

(v) माना * : N × N → N इस प्रकार परिभाषित है कि a * b = a तथा b का L.C.M.
∴ a = 1, b = 1 के लिए,
a * b = 1 = b * a
अत: 1a* संक्रिया के लिए व्युत्क्रमणीय है।

The step by step instructions on how to find the least common multiple of 12 and 16.

प्रश्न 7.
क्या समुच्चय {1, 2, 3, 4, 5} में a * b = a तथा b का LCM द्वारा परिभाषित * एक द्विआधारी संक्रिया है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
दिया गया समुच्चय = {1, 2, 3, 4, 5} द्विआधारी संक्रिया द्वारा परिभाषित है कि a * b = a और b का LCM 2 * 6 = 6 जो कि समुच्चय {1, 2, 3, 4, 5} में नहीं है इसलिए * एक द्विआधारी संक्रिया है।

प्रश्न 8.
मान लीजिए कि N में a * b = a तथा b का HCF द्वारा परिभाषित एक द्विआधारी संक्रिया है। क्या * क्रमविनिमेय है? क्या * साहचर्य है? क्या N में इस द्विआधारी संक्रिया के तत्समक का अस्तित्व है?
हल:
यहाँ N, प्राकृत संख्याओं का समुच्चय है।
द्विआधारी संक्रिया a * b = a, b का H.C.F. द्वारा परिभाषित
(i) a, b का H.C.F. = b, a के H.C.F.
a * b = b * a
अतः संक्रिया क्रमविनिमेय है।

(ii) a * (b * c)= a * (b, c का H.C.F.)
=a व b, c का H.C.F.
= a, b, c का H.C.F.
(a * b) * c = (a, b का H.C.F.) * c
= a, b व c का H.C.F.
= a, b, c का H.C.F.
a * (b * c)= (a * b) * c (∵ संक्रिया साहचर्य है)

(iii) 1 * a = a * 1 = 1 ≠ a
अतः तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 9.
मान लीजिए कि परिमेय संख्याओं के समुच्चय में निम्नलिखित प्रकार से परिभाषित * एक द्विआधारी संक्रिया है:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = \(\frac{a b}{4}\)
(vi) a * b = ab²
ज्ञात कीजिए कि इनमें से कौन-सी संक्रियाएँ क्रमविनिमेय हैं और कौन-सी साहचर्य हैं।
हल:
यहाँ परिमेय संख्याओं का समुच्चय Q दिया है।
(i) a * b = ab – b, द्विआधारी संक्रिया है।
(a) b * a = b – a
∴ a – b ≠ b – a ⇒ a * b ≠ b * a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c
(a * b) * c = (a-b)* c = a – b -c
∴ a – b + c ≠ a – b – c = a * (b * c) * (a * b) * c
अतः यह संक्रिया साहचर्य नहीं है।

(ii) (a) a * b = a² + b²
∴ b * a = b² + a² = a² + b²
⇒ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
(a + b) * c = (a² + b²) * c = (a² + b²)² + c²
⇒ a * (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

(iii) संक्रिया a * b = a + ab द्वारा परिभाषित है।
(a) a * b = a (1 + b), b * a = b + ba = b (1 + a)
∴ a * b + b * a
अतः यह * संक्रिया क्रमविनिमेय नहीं है।
(a) a * (b * c) = a + (b + bc)= a + a (b + bc)
= a + ab + abc
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c
= a + ab + ac + abc
∴ a * (b * c) (ab) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(iv) दिया है : a * b = (a – b)²
(a) a * b = (a – b)², b * a = (b – a)² = (a – b)²
∴ a * b = b * a
अतः यह * संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b – c) = [a – (b – c)²]²
(a * b) * c = (a – b)² * c = [(a – b)² – c]²
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(v) a * b = \(\frac{a b}{4}\)

अतः यह * संक्रिया साहचर्य है।

(vi) a * b = ab²
(a) a * b = ab², b * a = ba
∴ a * b ≠ b * a
अत: यह * संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a + bc² = a (bc²)² = ab²c⁴
(a * b) * c = ab² * c = ab²c² = ab²c²
∴ a + (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

प्रश्न 10.
सिद्ध कीजिए कि प्रश्न 9 में दी गई संक्रियाओं में किसी का तत्समक है, वह बतलाइए।
हल:
यहाँ (i) a * b = a – b
यदि e तत्समक अवयव हो तो
a * e = a – e, e * a = e – a
∴ a – e ≠ e – a ⇒ a * e ≠ e * a
अतः e का अस्तित्व नहीं है।

(ii) a * b = a² + b²
∴ a * e = a² + e², e * a = e² + a²
a * e = e * a ≠ a
अतः e का अस्तित्व नहीं है।

(iii) a * b = a + ab
a * e = a + ae, e * a = e + ea
∴ a * e # e * a # a
अत: e का अस्तित्व नहीं है।

(iv) a * b = (a – b)
a * e = (a – e)² # a, e * a = (e – a)² # a
a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(v) a * b = \(\frac{a b}{4}\)
a * e = \(\frac{ae}{4}\) # a, e * a = \(\frac{ea}{4}\) # a
∴ a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(vi) a * b = ab²
a * e = ae² # a, e * a = ea² # a
∴ a * e # e * a # a
अतः e का अस्तित्व नहीं है।

प्रश्न 11.
मान लीजिए कि A = N × N है तथा A में (a, b) * (c, d) = (a + c, b + d)द्वारा परिभाषित एक द्विआधारी संक्रिया है। सिद्ध कीजिए कि * क्रम विनिमेय तथा साहचर्य है। A में * का तत्समक अवयव, यदि कोई है, तो ज्ञात कीजिए।
हल:
माना A = N × N
द्विआधारी संक्रिया (Binary operation) * इस प्रकार परिभाषित है कि
(a, b) * (c, d) = (a + c, b + d)
इसलिए (c, d) * (a, b) = (c + a, d + b)
=(a + c, b + d)
=(a, b) * (c, d)
अतः द्विआधारी संक्रिया * क्रम विनिमेय है
पुनः (a, b)* [(c, d) * (e, f)]
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
तथा [(a, b) * (c, d)] * (e, f)
= (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)
= (a, b) * [(c, d) * (e, f)]
= [(a, b) * (c, d)]* (e, f)
अतः दी गई संक्रिया * साहचर्य है।
A में तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 12.
बतलाइए कि क्या निम्नलिखित कथन सत्य हैं या असत्य हैं। औचित्य भी बतलाइए।
(i) समुच्चय N में किसी भी स्वेच्छ द्विआधारी संक्रिया* के लिए a * a = a, ∀a ϵ N
(ii) यदि N में * एक क्रमविनिमेय द्विआधारी संक्रिया है तो a * (b * c)=(c * b) * a
हल:
यहाँ द्विआधारी संक्रिया समुच्चय N पर इस प्रकार परिभाषित की गयी है कि
a * a = a ∀ a ϵ N
(i) यहाँ पर * संक्रिया में केवल एक ही अवयव का प्रयोग किया गया है।
अतः यह कथन असत्य है।
(ii) वास्तविक संख्याओं में समुच्चय पर संक्रिया क्रमविनिमेय है।
b * c = c * b
= (c * b) * a = (b * c) * a = a * (b * c)
∴ a * (b * c) = (c * b) * a
अतः यह कथन सत्य है।

प्रश्न 13.
a * b = a³ + b³ प्रकार से परिभाषित N में एक द्विआधारी संक्रिया * पर विचार कीजिए। अब निम्नलिखित में से सही उत्तर का चयन कीजिए।
(A) * साहचर्य तथा क्रमविनिमेय दोनों है
(B) * क्रमविनिमेय है किन्तु साहचर्य नहीं है
(C) * साहचर्य है किन्तु क्रमविनिमेय नहीं है
(D) * न तो क्रमविनिमेय है और न साहचर्य है
हल:
यहाँ द्विआधारी संक्रिया को समुच्चय पर इस प्रकार परिभाषित किया गया है कि
a * b = a³ + b³
(a) a * b = a³ + b³, b * a = b³ + a³ = a³ * b³
∴ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b³ + c³) = a³ + (b³ + c³)³
(a * b) * c= (a³ + b³) * c = (a³ + b³) + c³
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।
∴ संक्रिया क्रमविनिमेय परन्तु साहचर्य नहीं है।
अतः विकल्प (B) सही है।

MP Board Class 6th Maths Solutions Chapter 8 Decimals Ex 8.1

maths decimal rounding to the nearest hundredth ·

Question 1.
Write the following as numbers in the given table.

Solution:

When you round to 1 decimal place, you inspect the hundredths integer.

Question 2.
Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Solution:

What is 1/8 as a decimal? You can also take the fraction 3/4 and write it in highest terms to 75/100.

Question 3.
Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solution:
(a) Seven-tenths = 7 tenths = \(\frac{7}{10}\) = 0.7
(b) Two tens and nine-tenths = \(2 \times 10+\frac{9}{10}\)
= 20 + 0.9 = 20.9
(c) Fourteen point six = 14.6
(d) One hundred and two-ones = 100 + 2 × 1
= 100 + 2 = 102
(e) Six hundred point eight = 600.8

Question 4.
Write each of the following as decimals:

Solution:

To write 9/20 as a decimal you have to divide numerator by the denominator of the fraction.

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form.
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solution:

Question 6.
Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Solution:

Question 7.
Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numbers is nearer the number?

Solution:
(a) 0.8
(b) 5.1
(C) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solution:
(a) 0.8 lies between 0 and 1. From 0 to 1, 0.8 is nearer to 1
(b) 5.1 lies between 5 and 6. From 5 to 6, 5.1 is nearer to 5.
(c) 2.6 lies between 2 and 3. From 2 to 3, 2.6 is nearer to 3.
(d) 6.4 lies between 6 and 7. From 6 to 7, 6.4 is nearer to 6.
(e) 9.1 lies between 9 and 10. From 9 to 10, 9.1 is nearer to 9.
(f) 4.9 lies between 4 and 5. From 4 to 5, 4.9 is nearer to 5.

Question 8.
Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solution:

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line.

Solution:

Question 10.
(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solution:
(a) 9 cm 5 mm = 9 cm + 5 mm
= \(\left(9+\frac{5}{10}\right)\) cm = 9.5 cm

(b) 65 mm = \(\frac{65}{10}\) cm = 6.5 cm

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.4

The factors of 84 are 1, 2, 3, 6, 7, 14, 21, and 42.

Question 1.
Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120
Solution:
(a) Factors of 20 are 1, 2, 4, 5, 10 and 20
Factors of 28 are 1, 2, 4, 7,14 and 28
∴ Common factors of 20 and 28 are 1, 2 and 4

(b) Factors of 15 are 1, 3, 5 and 15
Factors of 25 are 1, 5 and 25
∴ Common factors of 15 and 25 are 1 and 5

(c) Factors of 35 are 1, 5, 7 and 35
Factors of 50 are 1, 2, 5, 10, 25 and 50
∴ Common factors of 35 and 50 are 1 and 5

(d) Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56
Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120
∴ Common factors of 56 and 120 are 1, 2, 4 and 8

What is the greatest common factor of 12 and 18 ? 6.

Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5,15 and 25
Solution:
(a) Factors of 4 are 1, 2 and 4 Factors of 8 are 1, 2, 4 and 8 Factors of 12 are 1, 2, 3, 4, 6 and 12
∴ Common factors of 4, 8 and 12 are 1, 2 and 4

(b) Factors of 5 are 1 and 5 Factors of 15 are 1, 3, 5 and 15
Factors of 25 are 1, 5 and 25 .-. Common factors of 5, 15 and 25 are 1 and 5

Find the factors of 18 and 40.

Question 3.
Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Solution:
(a) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72,
∴ First three common multiples of 6 and 8 are 24, 48 and 72

(b) Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120,
Multiples of 18 are 18, 36, 54, 72, 90, 108, 126,
∴ First three common multiples of 12 and 18 are 36, 72 and 108

Question 4.
Write ail the numbers less than 100 which are common multiples of 3 and 4.
Solution:
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99,
Multiples of 4 are 4, 8, 12,16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100
Common multiples of 3 and 4 which are less than 100 are 12, 24, 36, 48, 60, 72, 84 and 96

Question 5.
Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16
Solution:
(a) Factors of 18 are 1, 2, 3, 6, 9 and 18
Factors of 35 are 1, 5, 7 and 35 Common factor of 18 and 35 is 1 Since, both have only one common factor, i.e., 1.
Therefore, 18 and 35 are co-prime numbers.

(b) Factors of 15 are 1, 3, 5 and 15
Factors of 37 are 1 and 37 Common factor of 15 and 37 is 1
Since, both have only one common factor, i.e., 1. Therefore, 15 and 37 are co-prime numbers.

(c) Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30
Factors of 415 are 1, 5, , 83 and 415
Common factors of 30 and 415 are 1 and 5
Since, both have more than one common factor. Therefore, 30 and 415 are not co-prime numbers.

(d) Factors of 17 are 1 and 17
Factors of 68 are 1, 2, 4, 17, 34 and 68 Common factors of 17 and 68 are 1 and 17
Since, both have more than one common factor. Therefore, 17 and 68 are not co-prime numbers.

(e) Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216
Factors of 215 are 1, 5, 43 and 215 Common factor of 216 and 215 is 1 Since, both have only one common factor, i.c., 1. Therefore, 216 and 215 are co-prime numbers.

(f) Factors of 81 are 1, 3, 9, 27 and 81 Factors of 16 are 1, 2, 4, 8 and 16 Common factor of 81 and 16 is 1 Since, both have only one common factor, i.e., 1. Therefore, 81 and 16 are co-prime numbers.

The prime factors of 50 are the prime numbers which divide 50 exactly, without remainder as defined by the Euclidean division.

Question 6.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
Since 5 × 12 = 60. The number divisible by both 5 and 12, must also be divisible by 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution:
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Therefore, the number divisible by 12, will also be divisible by 1, 2, 3, 4 and 6.

MP Board Class 6th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.1

The prime factorization of 84 is 2 x 2 x 3 x 7 x 1 = 22 x 3 x 7 x 1.

Question 1.
Find the common factors of the given terms,
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p²q²
(iv) 2x, 3x², 4
(v) 6abc, 24ab²,12a²b
(vi) 16x³, – 4x², 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x²y³, 10x³y², 6x²y²z
Solution:
(i) The given terms are 12x and 36.
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
Thus, the common factor of the given terms is 2 × 2 × 3 = 12

(ii) The given terms are 2y and 22xy
2y = 2 xy
22xy = 2 × 11 × X × y
Thus, the common factor of the given terms is 2 × y = 2y.

(iii) The given terms are 14pq and 18p²q²
14 pq = 2 × 7 × p × q
18p2q2 = 2 × 2 × 7 × p × p × q × q
Thus, the common factors of the given terms is 2 × 7 × p × q = 14pq

(iv) The given terms are 2x, 3x² and 4
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
Hence, the given three terms have no factor in common except 1.

(v) The given terms are 6abc, 24ab² and 12a²b
6abc= 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b.
12 a²b = 2 × 2 × 3 × a × a × b
Thus, the common factor of the given terms is 2 × 3 × a × b = 6ab.

(vi) The given terms are 16x³, -4x² and 32x
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = -1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
Thus, the common factor of the given terms is 2 × 2 × x = 4x.

(vii) The given terms are 10pq, 20qr and 30rp
10pq = 2 × 5 × p × q
20 qr = 2 × 2 × 5 × q × r
30 rp.= 2 × 3 × 5 × r × p
Thus, the common factor of the given terms is 2 × 5 = 10.

(viii)The given terms are 3x²y³, 10x³y² and 6x²y²z
3x²y² = 3 × x × x × y × y × y
10x³y³ = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × Z
Thus, the common factor of the given terms is x × x y × y = x²y².

Factors of 70 are any integer that can be multiplied by another integer to make exactly 70.

Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12g
(iii) 7a² + 14a
(iv) -16z + 20x³
(v) 20l²m + 30alm
(vi) 5x²y – 15xy²
(vii) 10a² – 15b² + 20c²
(viii) – 4a² + 4ab – 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
Solution:
(i) The expression is 7x – 42
Factors of 7x = 7 × x and 42 = 2 × 3 × 7
∴ 7x – 42 = 7 × x – 2 × 3 × 7 = 7(x – 2 × 3) = 7(x – 6)

(ii) The expression is 6p – 12q
Factors of 6p = 2 × 3 × p and
12q = 2 × 2 × 3 × q
∴ 6p – 12 q = 2 × 3 × p – 2 × 2 × 3 × q
= 2 × 3(p – 2 × q) = 6(p – 2 q)

(iii) The expression is 7a² + 14a
Factors of 7a² = 7 × a × a and
14a = 2 × 7 × a
∴ 7a² + 14a = 7 × a × a + 2 × 7 × a
= 7 × a(a + 2) = 7a (a + 2)

(iv) The expression is -16z + 20z³
Factors of -16z = – 1 × 2 × 2 × 2 × 2 × z and
20z³ = 2 × 2 × 5 × z × z × z
∴ -16z + 20z³ = -1 × 2 × 2 × 2 × 2 × z + 2 × 2 × 5 × z × z × z
= 2 × 2 × z(-1 × 2 × 2 + 5 × z × z)
= 4z (- 4 + 5z²)

(v) The expression is 20l²m + 30alm
Factors of 20l²m = 2 × 2 × 5 × l × l × m and
30alm = 2 × 3 × 5 × a × l × m;
∴ 20l²m + 30alm = 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m
= 2 × 5 × l × m (2 × l + 3 × a)
= 10lm (2l + 3a)

(vi) The expression is 5x²y – 15xy²
Factors of 5x²y = 5 × x × x × y and
15xy² = 3 × 5 × x × y × y
∴ 5x²y – 15xy² = 5 × x × x × y – 3 × 5 × x × y × y
= 5 × x × y (x – 3 × y) = 5xy (x – 3y).

(vii) The expression is 10a² – 15b² + 20c²
Factors of 10a² = 2 × 5 × a × a
-15b² = (-1) × 3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
∴ 10a² – 15b² + 20c²
= 2 × 5 × a × a – 3 × 5 × b × b + 2 × 2 × 5 × c × c
= 5(2a² – 3b² + 4c²)

(viii) The expression is – 4a² + 4ab – 4ca
Factors of – 4a² = (-1) × 2 × 2 × a × a
4ab = 2 × 2 × a × b and
-4ca = -1 × 2 × 2 × c × a
∴ -4a² + 4ab – 4ca = -1 × 2 × 2 × a × a + 2 × 2 × a × b – 1 × 2 × 2 × c × a
= 2 × 2 × a (-1 × a + b – 1 × c)
= 4a(-a + b – c)

(ix) The expression is x²yz + xy²z + xyz²
Factors of x²yz = x × x × y × z and
xy²z = x × y × y × z and xyz² = x × y × z × z
∴ x² yz + xy²z + xyz² = x × x × y × z + x × y × y × z + x × y × z × z
= x × y × z(x + y + z) = xyz (x + y + z)

(x) The expression is ax²y + bxy² + cxyz
Factors of ax²y = a × x × x × y,
bxy² = b × x × y × y and cxyz = c × x × y × z
∴ ax²y + bxy² + cxyz = a × x × x × y + b × x × y × y + c × x × y × z
= xy (a × x + b × y + c × z) = xy (ax + by + cz)

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. There are 8 integers that are factors of 24.

Question 3.
Factorise.
(i) x² + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay-by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution:
(i) The given expression is
x² + xy + 8x + 8y
= x × x + x × y + 8 × x + 8xy
= x(x + y) + 8(x + y) = (x + 8) (x + y).

(ii) The given expression is 15xy – 6x + 5y – 2
= 3x(5y – 2) + 1 × (5y – 2)
= (3x + 1) (5y – 2)

(iii) The given expression is ax + bx – ay – by
= x (a + b) – y (a + b) = (x – y) (a + b)

(iv) The given expression is
15 pq + 15 + 9 q + 25 p
= 15pq + 25p + 15 + 9q = 5p(3q + 5) + 3(5 + 3q)
= (5p + 3) (3q + 5)

(v) The given expression is z – 7 + 7xy – xyz
= z – 7 + 7 × x × y – x × y × z
= z – 7 + xy (7 – z) = 1 (z – 7) – xy (z – 7)
= (z – 7)(1 – xy)

Factors of 144 are any whole number that can be multiplied by another whole number to make exactly 144.

MP Board Class 8th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Solution:
(a) The sum of any two odd numbers is an even number.
As like, 1 + 3 = 4, 3 + 5 = 8

(b) The sum of any two even numbers is an even number.
As like, 2 + 4 = 6, 6 + 8 = 14

The factor pair of 96 or any number as a set of two factors, which, when multiplied together, give a particular product.

Question 2.
State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution:
(a) False
Since, sum of two odd numbers is even and sum of one odd number and one even number is always odd.
(b) True
Since, sum of two odd numbers is even and sum of two even numbers is always even.
(c) True
(d) False
If an even number is divided by 2, then the quotient is either odd or even.
(e) False
Since, prime number 2 is even.
(f) False
Factors of prime numbers are 1 and the number itself.
(g) False
Sum of two prime numbers is either even or odd.
(h) True
(i) False
Since, even number 2 is prime i.e., not composite.
(j) True

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
Pairs of prime numbers having same digits upto 100 are 17 and 71; 37 and 73; 79 and 97

So when we talk aqbout prime factorization of 38, we’re talking about the building blocks of the number.

Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

GCF Calculator is a free online tool that displays the Greatest Common Factor of two numbers given the inputs with detailed steps on how to approach.

Question 5.
What is the greatest prime number between 1 and 10?
Solution:
The greatest prime number between 1 and 10 is 7.

Question 6.
Express the following as the sum of two odd primes.
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 3 + 41
(b) 36 = 5 + 31
(c) 24 = 7 + 17
(d) 18 = 7 + 11

Question 7.
Give, three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes].
Solution:
Three pairs of prime numbers whose difference is 2 are 3 and 5; 5 and 7; 11 and 13.

Question 8.
Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26
Solution:
23 and 37 are prime numbers and 51 and 26 are composite numbers.
Thus, numbers in option (a) and (c) are prime.

Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
Seven consecutive composite numbers less than 100 are 90, 91, 92, 93, 94, 95, 96

Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution:
(a) 21 = 3 + 7 + 11
(b) 31 = 3 + 11 + 17
(c) 53 = 13 + 17 + 23
(d) 61 = 13 + 19 + 29

Question 11.
Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
(Hint :3 + 7 = 10)
Solution:
Since, 2 + 3 = 5; 7 + 13 = 20; 3 + 17 = 20; 2 + 13 = 15; 5 + 5 = 10 and 5, 10, 15, 20 all are
divisible by 5.
So, five pairs of prime numbers less than 20 whose sum is divisible by 5 are 2, 3; 2, 13; 3, 17; 7, 13; 5, 5.

Question 12.
Fill in the blanks:
(a) A number which has only two factors is called a ___.
(b) A number which has more than two factors is called a ___.
(c) 1 is neither ___ nor ___.
(d) The smallest prime number is ___.
(e) The smallest composite number is ___.
(f) The smallest even number is ___.
Solution:
(a) Prime number
(b) Composite number
(c) Prime number, composite number
(d) 2
(e) 4
(f) 2

MP Board Class 6th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 16 Playing with Numbers Ex 16.2

Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution:
The given number is 21y5
For this number to be multiple of 9, the sum of its digits should be a multiple of 9.
∴ 2 + 1 + y + 5 is a multiple of 9.
or 8 + y is a multiple of 9.
For y to be a single digit number,
8 + y = 9 ⇒ y = 1

Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution:
The given number is 31z5.
For this number to be multiple of 9, the sum of its digits should be a multiple of 9.
∴ 3 + 1 + z + 5 is a multiple of 9.
or z + 9 is a multiple of 9.
∴ For z to be a single digit number,
z + 9 = 9 or z + 9 = 18
⇒ z = 0 or z = 9
Hence, z = 0, 9
We get two answers, because 9 and 18, both are multiples of 9.

The Sum and difference of Cubes Calculator is a free online tool.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x? (Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers :0, 3, 6, 9, 12, 15, 18, But since x is a digit, it can only be that 6 + x = 6 or 9 o r12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values).
Solution:
The given number 24x is a multiple of 3.
⇒ Sum of the digits of this number are also a multiple of 3.
∴ 2 + 4 + x = x + 6 is a multiple of 3.
For x to be a single digit number,
x + 6 = 6, 9, 12, 15 or x = 0, 3, 6, 9.

Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
As the given number 31z5 is a multiple of 3, the sum of its digits should also be a multiple of 3.
∴ 3 + 1 + z + 5 is a multiple of 3.
or z + 9 is a multiple of 3.
Now, for z to be a single digit number,
z + 9 = 9, 12, 15, 18 or z = 0, 3, 6, 9

MP Board Class 8th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find:

Solution:

A mixed number is an Convert to 2 2/3 Improper Fraction.

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

Solution:
(i) Reciprocal of \(\frac{3}{7}=\frac{7}{3}\). It is an improper fraction.
(ii) Reciprocal of \(\frac{5}{8}=\frac{8}{5}\). It is an improper fraction.
(iii) Reciprocal of \(\frac{9}{7}=\frac{7}{9}\). It is a proper fraction.
(iv) Reciprocal of \(\frac{6}{5}=\frac{5}{6}\). It is a proper fraction.
(v) Reciprocal of \(\frac{12}{7}=\frac{7}{12}\). It is a proper fraction.
(vi) Reciprocal of \(\frac{1}{8}\) = 8. It is a whole fraction.
(vii) Reciprocal of \(\frac{1}{11}\) = 11. It is a whole fraction.

Question 3.

Question 4.
Find:

Solution:
(i) \(\frac{2}{5}÷\frac{1}{2}=\frac{2}{5} \times 2=\frac{4}{5}\)

MP Board Class 7th Maths Solutions