Quantitative Aptitude - Arithmetic Ability Questions

Let Meena’s age = A.
Then Sivagami’s age = A + 2
After 6 years the total of their ages will be 7 times of what?
Not clear.
So the given data are inadequate.

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 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 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 5C2*3C2= 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 5C3 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 5C3 = 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.

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

 512=K1-154    ∵ 20% =20100=15

        or      

512=K454

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

   -----------(1)

                --------------(2)

(1) - (2) gives

 th of temperature on Monday = 

 Temperature on Monday = 

 Temperature on Thursday = 

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 %.