6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?

A) 36	B) 25
C) 24	D) 72

Answer & Explanation Answer: B) 25

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1

b. 3, 1, 1

Case a. Number of ways of achieving the first option 2 - 2 - 1

Two toys out of the 5 can be selected in $5 C_{2}$ ways. Another 2 out of the remaining 3 can be selected in $3 C_{2}$ ways and the last toy can be selected in $1 C_{1}$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways $5 C_{2} * 3 C_{2}$ = 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in $5 C_{3}$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is $5 C_{3}$ = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

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Filed Under: Permutations and Combinations

34 25852

Q:

Find the odd figure?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: B) 2

Explanation:

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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 25806

Q:

A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar, the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was filled with the sugar solution. The initial amount of honey in the jar was:

A) 1.25 kg	B) 1 kg
C) 1.5 kg	D) None of these

Answer & Explanation Answer: A) 1.25 kg

Explanation:

Let the initial amount of honey in the jar was K, then

$512 = K {(1 - \frac{1}{5})}^{4}$ $[∵ 20 % = \frac{20}{100} = \frac{1}{5}]$

or

$512 = K {(\frac{4}{5})}^{4}$

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

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Filed Under: Alligation or Mixture

32 25796

Q:

Select the odd number from the given alternatives.

A) 25	B) 49
C) 9	D) 8

Answer & Explanation Answer: D) 8

Explanation:

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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 25794

Q:

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days ?

A) 9 days	B) 11 days
C) 13 days	D) 15 days

Answer & Explanation Answer: C) 13 days

Explanation:

Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x => x = ( 23 x 13/10 ) => x = 299 /10.

A's 1 day's work = 1/23 ;
B's 1 day's work = 10/299 .

(A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 .

Therefore, A and B together can complete the work in 13 days.

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Filed Under: Time and Work
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

52 25763

Q:

The mean temperature of Monday to Wednesday was 37C and of Tuesday to Thursday was 34C . If the temperature on Thursday was (4/5) th that of Monday, the temperature on Thursday was?

Answer

$\inline \fn_jvn M+T+W =(37\times 3)^{0}=111^{0}C$ -----------(1)

$\inline \fn_jvn T+W+Th=(34\times 3)^{0}=102^{0}C$

$\inline \fn_jvn \Rightarrow T+W+\frac{4}{5}M = 102^{0}C$ --------------(2)

(1) - (2) gives

$\inline \fn_jvn \frac{1}{5}$ th of temperature on Monday = $111^{0}-102^{0}=9^{0}C$

$\inline \fn_jvn \Rightarrow$ Temperature on Monday = $\inline \fn_jvn 45^{0}C$

$\inline \fn_jvn \therefore$ Temperature on Thursday = $\inline \fn_jvn \frac{4}{5}\times 45^{0}=36^{0}$

Workspace

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Subject: Average

160 25740

Q:

A bag contains 600 coins of 25 p denomination and 1200 coins of 50 p denomination. If 12% of 25 p coins and 24% of 50 p coins are removed, the percentage of money removed from the bag is nearly :

A) 21.6 %	B) 15.3 %
C) 14.6 %	D) 12.5 %

Answer & Explanation Answer: A) 21.6 %

Explanation:

Total money = Rs.[600*(25/100)+1200*(50/100)]= Rs. 750.

25 paise coins removed = Rs. (600*12/100) = 72.

50 paise coins removed = Rs. (1200*24/100)= 288.

Money removed =Rs.(72*25/100+288*50/100) = Rs.162.

Required percentage = (162/750*100)% = 21.6 %.

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Filed Under: Percentage

57 25737

Q:

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

A) 32	B) 24
C) 72	D) 36

Answer & Explanation Answer: A) 32

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

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Filed Under: Permutations and Combinations

17 25684

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