Quantitative Aptitude - Arithmetic Ability Questions

Given expression = 0.2×0.2×0.010.1×0.1×0.02=0.04+0.010.01+0.02=0.050.03=53

Let the smallest number be x.
Then larger number = (x + 1365)
x + 1365 = 6x + 15
= 5x = 1350
x = 270
Smaller number = 270.

Let the required height of the building be x meter

More shadow length, More height(direct proportion)

Hence we can write as

(shadow length) 40.25 : 28.75 :: 17.5 : x
⇒ 40.25 × x = 28.75 × 17.5
⇒ x = 28.75 × 17.5/40.25
= 2875 × 175/40250
= 2875 × 7/1610
= 2875/230
= 575/46
= 12.5 cm

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

Since the compound interest is taken here,

 PW1+51002=1764

=> PW = 1600

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

From the given data,

K's profit : L's profit = 7x x 11 : 9x x 7

= 11:9

Therefore, L's profit = 9 x 2460/20 =  Rs. 1107

Watson bought the book for Rs. 240 and sold to Johny at a profit of 50%.

S.P = C.P(1 + P%/100)

=> S.P for Watson = C.P for Johny = 240(1 + 50/100) = 240 x 1.5 = Rs. 360

Let Johny quoted the marked price of the book as Rs. M

We know, SP = M.P(1 - Discount(%)/100)

Here discount = 10% to Shekar,

S.P for Johny = M(1 - 10/100) = 0.9M

But Johny want to earn 25% profit,

=> S.P = C.P(1 + P%/100)

=> 0.9M = 360(1 + 25/100)

=> M = (360x1.25)/0.9

=> M = Rs. 500

Therefore, Johny should quote Rs. 500 as the marked price of the book to get 25% profit and allowing 10% discount to Shekar.