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Q:

I. A, B, C, D, E and F are six members of a family.

II. One couple has parents and their children in the family.

III. A is the son of C and E is the daughter of A.

IV. D is the daughter of F who is the mother of E.

Who are the male members in the family ?

A) A and C	B) C and F
C) A, B and D	D) Cannot be determined

Answer & Explanation Answer: D) Cannot be determined

Explanation:

E is the daughter of A and F is the mother of E. So, A is the father of E and hence the husband of F. Now,

D is the daughter of F. So, D and E are the daughters of A and F Also. A is the son of C.

Now, only B remains. Thus, B and C are the parents of A.

The sex of B and C cannot be determined:

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Filed Under: Blood Relations

Q:

The ratios of areas of two squares, one having its diagonal double than the other is

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

Answer & Explanation Answer: D) 4:1

Explanation:

ratio = $\frac{(1 / 2) \times {(2 d)}^{2}}{\frac{1}{2} d^{2}} = 4 : 1$

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

Q:

The area of a square field is 69696sq.cm..its diagonal will be equal to

A) 373	B) 573
C) 473	D) 673

Answer & Explanation Answer: A) 373

Explanation:

$d = \sqrt{2 * s} = > d = \sqrt{2 * 69696}$

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Filed Under: Area
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Q:

If the length of the diagonal of a square is 20cm,then its perimeter must be

a. $40 \sqrt{2}$ cm b. $30 \sqrt{2}$ cm c. 10cm d. 15 $\sqrt{2}$ cm

A) a	B) b
C) c	D) d

Answer & Explanation Answer: A) a

Explanation:

We know that d=√2s

Given diagonal = 20 cm

=> s = 20/ $\sqrt{2}$ cm

Therefore, perimeter of the square is 4s = 4 x 20/ $\sqrt{2}$ = 40 $\sqrt{2}$ cm.

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Filed Under: Area
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Q:

The area of the circle is 220 sq.m, the area of the square inscribed in a circle will be

A) 49	B) 70
C) 140	D) 150

Answer & Explanation Answer: C) 140

Explanation:

area of a square = $\frac{1}{2} {(d i a g o n a l)}^{2}$

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Q:

The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is

A) 49	B) 154
C) 378	D) 1078

Answer & Explanation Answer: B) 154

Explanation:

The diameter is equal to the shortest side of the rectangle.

So radius= 14/2 = 7cm.

Therefore, $a r e a o f c i r c l e = {πr}^{2} = \frac{22}{7} \times 49 = 154 {cm}^{2}$

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Q:

At what time between 5.30 and 6 will the hands of a clock be at right angles?

A) (43 + 5/11) min past 5	B) (43 + 7/11) min pats 5
C) 5	D) 7

Answer & Explanation Answer: B) (43 + 7/11) min pats 5

Explanation:

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

55 min. spaces are gained in 60 min

40 min. spaces are gained in $\frac{60}{55} \times 40$ min

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Filed Under: Clocks
Exam Prep: Bank Exams
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Q:

At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but,not together?

A) (5 + 5/11 )min past 7	B) (6 + 5/11) min past 7
C) 7	D) 8

Answer & Explanation Answer: A) (5 + 5/11 )min past 7

Explanation:

When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
$\therefore$ Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.

5 min spaces are gained in $(\frac{60}{55} * 5)$ min = $5 \frac{5}{11}$ min

so, Required time = $5 \frac{5}{11}$ min past 7

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

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