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Q:

For any two numbers m , n ; (m+n) : (m-n) : mn = 7: 1: 60 . Find the value of 1/m : 1/n

A) 4:3	B) 8:7
C) 3:4	D) 7:8

Answer & Explanation Answer: C) 3:4

Explanation:

$\frac{m + n}{m - n} = \frac{7 x}{x} \Rightarrow \frac{m}{n} = \frac{4 x}{3 x}$

Again $m n = 12 x^{2}$

and mn =60x

so, $60 x = 12 x^{2} \Rightarrow x = 5$

=> m= 20 and n= 15

Hence, $\frac{1}{m} : \frac{1}{n} = \frac{1}{20} : \frac{1}{15} = 3 : 4$

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Filed Under: Ratios and Proportions

Q:

A child has three different kinds of chocolates costing Rs.2, Rs.5, Rs.10. He spends total Rs. 120 on the chocolates. what is the minimum possible number of chocolates he can buy, if there must be atleast one chocolate of each kind?

A) 22	B) 19
C) 17	D) 15

Answer & Explanation Answer: C) 17

Explanation:

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

Thus, 2 x 5 + 5 x 2 =Rs.20

Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.

Thus minumum number of chocolates = 5 + 2 + 10 = 17

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Filed Under: Ratios and Proportions

Q:

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

A) 1677	B) 1683
C) 2523	D) 3363

Answer & Explanation Answer: B) 1683

Explanation:

L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2

Therefore, Required Number = (840 x 2+3)=1683

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Filed Under: HCF and LCM

Q:

The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is

A) 1008	B) 1015
C) 1022	D) 1032

Answer & Explanation Answer: B) 1015

Explanation:

Required Number = (L.C.M of 12, 16, 18,21,28)+7

= 1008 + 7

= 1015

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Filed Under: HCF and LCM

Q:

Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case

A) 4	B) 7
C) 9	D) 13

Answer & Explanation Answer: A) 4

Explanation:

Required Number = H.C.F of (91- 43), (183- 91) and (183-43)

= H.C.F of 48, 92, and 140 = 4

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Filed Under: HCF and LCM

Q:

The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A) 91	B) 910
C) 1001	D) 1911

Answer & Explanation Answer: A) 91

Explanation:

Required number of students = H.C.F of 1001 and 910 = 91

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Filed Under: HCF and LCM

Q:

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2	B) -1
C) 1	D) 2

Answer & Explanation Answer: A) -2

Explanation:

H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

Therefore, 3y-x = (3 x 7)-23 = -2

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Filed Under: HCF and LCM

Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77	B) 88
C) 99	D) 110

Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35

Now, co-primes with product 35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 77

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Filed Under: HCF and LCM

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