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Q:

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A) 27 and 23	B) 24 and 23
C) 25 and 23	D) 22 and 23

Answer & Explanation Answer: A) 27 and 23

Explanation:

Let the two parallel sides of the trapezium be a cm and b cm.

Then, a - b = 4

And, $\frac{1}{2} \times (a + b) \times 19 = 475 = > (a + b) = 50$

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm.

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.

A) 4.5	B) 3.5
C) 2.5	D) 6.5

Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = $a \frac{\sqrt{3}}{2}$ = $3 \sqrt{3} \times \frac{\sqrt{3}}{2}$ = 4.5

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Filed Under: Area
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Q:

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A) 18	B) 28
C) 38	D) 48

Answer & Explanation Answer: A) 18

Explanation:

$\sqrt{(l^{2} + b^{2})} = \sqrt{41}$ (or) $l^{2} + b^{2} = 41$

Also, $l b = 20$

${(l + b)}^{2} = l^{2 +} b^{2} + 2 l b$ = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

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Filed Under: Area
Exam Prep: CAT , Bank Exams , AIEEE
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Q:

The diagonal of the floor of a rectangular closet is 7 $\frac{1}{2}$ feet. The shorter side of the closet is 4 $\frac{1}{2}$ feet. What is the area of the closet in square feet?

A) 9	B) 18
C) 27	D) 36

Answer & Explanation Answer: C) 27

Explanation:

Other side = $\sqrt{d^{2} - s^{2}}$ = $\sqrt{(\frac{225}{4} - \frac{81}{4})}$ = $\sqrt{\frac{144}{4}}$ = 6 ft

Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

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Filed Under: Area
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

Q:

If $n C_{10} = n C_{12}$ then,find n.

A) 10	B) 12
C) 22	D) 24

Answer & Explanation Answer: C) 22

Explanation:

Using, ${}^{n}C_{r} = {}^{n}C_{n - r}$ we get

n – 10 = 12

or, n = 12 + 10 = 22

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Filed Under: Permutations and Combinations

Q:

if $\log_{10} 5 + \log (5 x + 1) = \log_{10} (x + 5) + 1, t h e n x i s e q u a l t o$

A) 1	B) 3
C) 5	D) 10

Answer & Explanation Answer: B) 3

Explanation:

$\log_{10} 5 + \log (5 x + 1) = \log_{10} (x + 5) + 1$

$\log_{10} [5 (5 x + 1)] = \log_{10} [10 (x + 5)]$

$5 (5 x + 1) = 10 (x + 5)$

5x+1=2x+10

3x=9

x=3

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Filed Under: Logarithms
Exam Prep: Bank Exams
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Q:

What annual payment will discharge a debt of Rs.7620 due in 3years at $16 \frac{2}{3}$ % per annum interest?

A) 5430	B) 4430
C) 3430	D) 2430

Answer & Explanation Answer: C) 3430

Explanation:

Let each installment be Rs.x. Then,

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.

$[\frac{x}{(1 + \frac{50}{3 x 100})}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{2}}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{3}}] = 7620$

$\frac{6 x}{7} + \frac{36 x}{49} + \frac{216 x}{343} = 7620$

$294 x + 252 x + 216 x = 7620 x 343$

=> x = 3430

Amount of each installment = Rs.3430

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Filed Under: Compound Interest
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Q:

If the compound interest on a certain sum at $16 \frac{2}{3}$ % to 3 years is Rs.1270, find the simple interest on the same sum at the same rate and f or the same period.

A) 1080	B) 1090
C) 1180	D) 1109

Answer & Explanation Answer: A) 1080

Explanation:

Let the sum be Rs. x. Then,

$C . I = [x * {(\frac{50}{300 * 100})}^{3} - x] = (\frac{343 x}{216} - x) = \frac{127 x}{216}$

$\Leftrightarrow \frac{127 x}{216} = 1270$

$x = 2160$

Thus, the sum is Rs. 2160

$S . I = R s . [2160 * \frac{50}{3} * 3 * \frac{1}{100}] = R s 1080$

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Filed Under: Compound Interest
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