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# IBPS PO 2017 Quantitative aptitude Practice Quiz 2

IBPS EXAM Guru
1 . x = ($4\times 5\over55....\infty$)x $4^5$

 A.   4 B.   5 C.   $\infty$ D.   0
2 . What is the area of the right-angled triangle?
I.$sec^2 \theta - tan^2 \theta = 1$
II.$cot(90^o - \theta)$ = $12\over13$

 A.   1 B.   2 C.   3 D.   4
3 . Solving the simultaneous equations, what will be the value of P?
I.$7\over2$P+9q=42
II.$3\over2$P + $27\over7$q = 18

 A.   1 B.   2 C.   3 D.   4
4 . Direction (Q. 4 to 8): Each of these questions has 2 number series marked as I and II. Study the pattern of series I and apply it to complete series II, whose Ist place is filled up for you.
$Q$. I. 2160 360 72 18 6
II. 420 a b c d
What is the value of b?

 A.   70 B.   84 C.   41 D.   14
5 . I. 3 19 93 375 1121
II. 0 a b c d
Which number must replaced?

 A.   1 B.   2 C.   3 D.   4
6 . I. 2 0 –6 –18
II. 8 a b c
can be represented by:

 A.   -2 B.   -20 C.   1 D.   -12
7 . I. 4 17 153 2448
II. 2 p q r
The number r will be:

 A.   1331 B.   1234 C.   1243 D.   3131
8 . I. 32 37 47 58 71 79
II. 23 m n o p q
What should come in place of o?

 A.   36 B.   94 C.   14 D.   63
9 . The number 7, expressed as a percentage is:

 A.   0.07 B.   0.7 C.   7 D.   700
10 . The sum of the series 3 – 3 3 + 9 – 9 3 + ... upto 8 terms is:

 A.   1 B.   2 C.   3 D.   4
1 .
 Answer : Option D Explanation : $5^{55}....\infty$ $\cong$ $\infty$, Thus, $4\times5\times45\over \infty$ $\cong$0
2 .
 Answer : Option D Explanation : From I: no information From II: tan $\theta$ = $12\over13$ base = 13, highest = 12 But, tan is only a ratio of base and heights, not necessarily true values.
3 .
 Answer : Option D Explanation : I x $3\over7$ = I.Thus, both equations are same To solve for 2 values, at least 2 different equations are needed.
4 .
 Answer : Option D Explanation : 2160 ÷ 6 = 360, 360 ÷ 5 = 72, 72 ÷ 4 = 18, 18 ÷ 3 = 6. Thus, 420 ÷ 6 ÷ 5 = 14
5 .
 Answer : Option C Explanation : 3 x 6 + 1 = 19, 19 x 5 – 2 = 93, 93 x 4 + 3 = 375, 375 x 3 – 4 = 1589, etc Thus, 0 x 6 + 1 = 01, 01 x 5 – 2 = 03, 03 x 4 + 3 = 15, 15 x 3 – 4 = 41
6 .
 Answer : Option D Explanation : $2 - 2^ 2 + 2 = 0, 0 - 3^ 2 + 3 = -6, -6 -4^ 2 + 4 = - 18, etc$ $Thus, 8 - 2^ 2 + 2 = 6, 6 - 3^ 2 + 3 = 0, 0 - 4^ 2 + 4 = -12, etc$
7 .
 Answer : Option A Explanation : The series is: $x 2^2+1, 3^2+2,4^2+3, etc$ Thus, r =1331
8 .
 Answer : Option D Explanation : The given series is : 32 + (3 + 2) = 37, 37 + (3 + 7) = 47, 47 + (4 + 7) = 58, etc Thus, m = 23 + (2 + 3) = 28, n = 28 + (2 + 8) = 38, O = 38 + (3 + 8) = 49.
9 .
 Answer : Option D Explanation :
10 .
 Answer : Option D Explanation : $S_n$(Sum of n terms) = $a(rn-1)\over r-1$ = $3[1-\sqrt{318-1}]\over \sqrt{3-1}$ =$3(80)\over 0.7$ $\cong$ 343