# Arithmetic questions for SSC Stenographer PDF

0
402

## Arithmetic questions for SSC Stenographer PDF

SSC Stenographer Arithmetic questions and answers pdf based on previous year question papers of SSC exam. Top 15 Very important questions  for Stenographer.

Question 1: Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.

a) 16

b) 36

c) -16

d) -36

Question 2: If 9/4th of 7/2 of a number is 126, then 7/2th of that number is …………..

a) 56

b) 284

c) 72

d) 26

Question 3: The 4th term of an arithmetic progression is 15, 15th term is -29, ﬁnd the 10th term?

a) -5

b) -13

c) -17

d) -9

Question 4: (91 + 92 + 93 + ……… +110) is equal to

a) 4020

b) 2010

c) 6030

d) 8040

Question 5: If 6/7th of 8/5th of a number is 192, then 3/4th of that number is .———–

a) 105

b) 77

c) 36

d) 80

Question 6: 40.36 – (9.347 – x ) – 29.02 = 3.68. Find x.

a) -56.353

b) 1.687

c) -17.007

d) 82.407

Question 7: What is the value of (81 + 82 + 83 + ……… +130)?

a) 5275

b) 10550

c) 15825

d) 21100

Question 8: In an arithmetic progression if 13 is the 3rd term, ­47 is the 13th term, then ­30 is which term?

a) 9

b) 10

c) 7

d) 8

Question 9: If 4/5th of 6/7th of a number is 216, then 8/9th of that number will be

a) 179

b) 280

c) 160

d) 269

Question 10: 199994 x 200006 = ?

a) 39999799964

b) 39999999864

c) 39999999954

d) 39999999964

Question 11: In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?

a) 10

b) 11

c) 9

d) 12

Question 12: In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?

a) 11

b) 8

c) 10

d) 7

Question 13: 2*[­0.3(1.3 + 3.7)] of 0.8 = ?

a) ­1.92

b) ­0.72

c) ­2.16

d) ­2.4

Question 14: Find the value of p, if 2x – 4, 4x + p and 6x – 12 are in arithmetic progression.

a) -9

b) -10

c) -11

d) -8

Question 15: If 7/8th of 5/4th of a number is 315, then 5/9th of that number is _____ .

a) 123

b) 81

c) 140

d) 160

SSC Stenographer Previous Papers

Terms in arithmetic progression : $(3x + p) , (x – 10) , (-x + 16)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$

=> $-2x -10 – p = -2x + 16 + 10$

=> $-p = 26 + 10 = 36$

=> $p = -36$

=> Ans – (D)

Let the number be $x$

According to ques,

=> $\frac{9}{4} \times \frac{7}{2} \times x = 126$

=> $\frac{63}{8} x = 126$

=> $x = \frac{126}{63} \times 8$

=> $x = 2 \times 8 = 16$

$\therefore (\frac{7}{2})^{th}$ of the number = $\frac{7}{2} \times 16$

= $7 \times 8 = 56$

=> Ans – (A)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

4th term, $A_4 = a + (4 – 1) d = 15$

=> $a + 3d = 15$ —————–(i)

Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(14d – 3d) = -29 – 15$

=> $d = \frac{-44}{11} = -4$

Substituting it in equation (i), => $a – 12 = 15$

=> $a = 15 + 12 = 27$

$\therefore$ 10th term, $A_{10} = a + (10 – 1)d$

= $27 + (9 \times -4) = 27 – 36 = -9$

=> Ans – (D)

Expression : (91 + 92 + 93 + ……… +110)

This is an arithmetic progression with first term, $a = 91$ , last term, $l = 110$ and common difference, $d = 1$

Let number of terms = $n$

Last term in an A.P. = $a + (n – 1)d = 110$

=> $91 + (n – 1)(1) = 110$

=> $n – 1 = 110 – 91 = 19$

=> $n = 19 + 1 = 20$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{20}{2} (91 + 110)$

= $10 \times 201 = 2010$

Let the number be $x$

According to ques,

=> $\frac{6}{7} \times \frac{8}{5} \times x = 192$

=> $\frac{48}{35} x = 192$

=> $x = \frac{192}{48} \times 35$

=> $x = 4 \times 35 = 140$

$\therefore (\frac{3}{4})^{th}$ of the number = $\frac{3}{4} \times 140$

= $3 \times 35 = 105$

=> Ans – (A)

Expression : 40.36 – (9.347 – x ) – 29.02 = 3.68

=> 40.36 – 9.347 + x = 3.68 + 29.02

=> 31.013 + x = 32.7

=> x = 32.7 – 31.013

=> x = 1.687

=> Ans – (B)

Expression : (81 + 82 + 83 + ……… +130)

This is an arithmetic progression with first term, $a = 81$ , last term, $l = 130$ and common difference, $d = 1$

Let number of terms = $n$

Last term in an A.P. = $a + (n – 1)d = 130$

=> $81 + (n – 1)(1) = 130$

=> $n – 1 = 130 – 81 = 49$

=> $n = 49 + 1 = 50$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{50}{2} (81 + 130)$

= $25 \times 211 = 5275$

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 13$

=> $a + 2d = 13$ —————–(i)

Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(12d – 2d) = 47 – 13 = 34$

=> $d = \frac{34}{10} = 3.4$

Substituting it in equation (i), => $a + 2 \times 3.4 = 13$

=> $a = 13 – 6.8 = 6.2$

Let $n^{th}$ term = 30

=> $a + (n – 1) d = 30$

=> $6.2 + (n – 1) (3.4) = 30$

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$

=> $(n – 1) = \frac{23.8}{3.4} = 7$

=> $n = 7 + 1 = 8$

Let the number be $x$

According to ques,

=> $\frac{4}{5} \times \frac{6}{7} \times x = 216$

=> $x = 216 \times \frac{35}{24} = 9 \times 35$

$\therefore$ 8/9th of the number = $\frac{8}{9} \times (35 \times 9)$

= $8 \times 35 = 280$

=> Ans – (B)

Expression :  199994 x 200006

= (200000 – 6) x (200000 + 6)

= $(200000)^2 – (6)^2$

= 40000000000 – 36 = 39999999964

=> Ans – (D)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 17$

=> $a + 2d = 17$ —————–(i)

Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(16d – 2d) = -25 – 17$

=> $d = \frac{-42}{14} = -3$

Substituting it in equation (i), => $a – 6 = 17$

=> $a = 17 + 6 = 23$

Let $n^{th}$ term = -1

=> $a + (n – 1) d = -1$

=> $23 + (n – 1) (-3) = -1$

=> $(n – 1) (-3) = -1 – 23 = -24$

=> $(n – 1) = \frac{-24}{-3} = 8$

=> $n = 8 + 1 = 9$

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

5th term, $A_5 = a + (5 – 1) d = 9$

=> $a + 4d = 9$ —————–(i)

Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(11d – 4d) = -26 – 9$

=> $d = \frac{-35}{7} = -5$

Substituting it in equation (i), => $a – 20 = 9$

=> $a = 9 + 20 = 29$

Let $n^{th}$ term = -6

=> $a + (n – 1) d = -6$

=> $29 + (n – 1) (-5) = -6$

=> $(n – 1) (-5) = -6 – 29 = -35$

=> $(n – 1) = \frac{-35}{-5} = 7$

=> $n = 7 + 1 = 8$

Expression : 2*[­0.3(1.3 + 3.7)] of 0.8 = ?

= $2 \times (0.3 \times 5) \times 0.8$

= $10 \times 0.24 = 2.4$

=> Ans – (D)

Terms in arithmetic progression : $(2x – 4) , (4x + p) , (6x – 12)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(4x + p) – (2x – 4) = (6x – 12) – (4x + p)$

=> $2x + p + 4 = 2x – 12 – p$

=> $2p = -12 – 4 = -16$

=> $p = \frac{-16}{2} = -8$

Let the number be $x$
=> $\frac{7}{8} \times \frac{5}{4} \times x = 315$
=> $x = 9 \times 8 \times 4 = 32 \times 9$
$\therefore$ 5/9th of the number = $\frac{5}{9} \times 32 \times 9$