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.
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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
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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
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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
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Answers & Solutions:
1) Answer (D)
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)
2) Answer (A)
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)
3) Answer (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.
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)
4) Answer (B)
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. = $\frac{n}{2} (a + l)$
= $\frac{20}{2} (91 + 110)$
= $10 \times 201 = 2010$
5) Answer (A)
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)
6) Answer (B)
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)
7) Answer (A)
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. = $\frac{n}{2} (a + l)$
= $\frac{50}{2} (81 + 130)$
= $25 \times 211 = 5275$
8) Answer (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, $A_3 = a + (3 – 1) d = 13$
=> $a + 2d = 13$ —————–(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
=> $a + (n – 1) d = 30$
=> $6.2 + (n – 1) (3.4) = 30$
=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$
=> $(n – 1) = \frac{23.8}{3.4} = 7$
=> $n = 7 + 1 = 8$
9) Answer (B)
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)
10) Answer (D)
Expression : 199994 x 200006
= (200000 – 6) x (200000 + 6)
= $(200000)^2 – (6)^2$
= 40000000000 – 36 = 39999999964
=> Ans – (D)
11) Answer (C)
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
=> $a + (n – 1) d = -1$
=> $23 + (n – 1) (-3) = -1$
=> $(n – 1) (-3) = -1 – 23 = -24$
=> $(n – 1) = \frac{-24}{-3} = 8$
=> $n = 8 + 1 = 9$
12) Answer (B)
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
=> $a + (n – 1) d = -6$
=> $29 + (n – 1) (-5) = -6$
=> $(n – 1) (-5) = -6 – 29 = -35$
=> $(n – 1) = \frac{-35}{-5} = 7$
=> $n = 7 + 1 = 8$
13) Answer (D)
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)
14) Answer (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
=> $(4x + p) – (2x – 4) = (6x – 12) – (4x + p)$
=> $2x + p + 4 = 2x – 12 – p$
=> $2p = -12 – 4 = -16$
=> $p = \frac{-16}{2} = -8$
15) Answer (D)
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 = $\frac{5}{9} \times 32 \times 9$
= 160
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