適性-基本的な算術の例

Q 1 - AP 5、8、11、14、17、...の^第16項は次のうちどれですか？

A -50

B -51

C -52

D -53

Answer - A

Explanation

Here a = 5, d = 8 - 5 = 3, n = 16
Using formula T_n = a + (n - 1)d
T₁₆ = 5 + (16 - 1) x 3
= 50

Q 2 -AP 4、9、14、19、24、...の次の用語のどれが109ですか？

A -20番目

B - 21日

C - 22 ND

D - 23日

Answer - C

Explanation

Here a = 4, d = 9 - 4 = 5
Using formula T_n = a + (n - 1)d
T_n = 4 + (n - 1) x 5 = 109 where 109 is the n^th term.
=> 4 + 5n - 5 = 109
=> 5n = 109 + 1 
=> n = 110 / 5 
= 22

Q 3 -AP 7、13、19、... 205にはいくつの用語がありますか？

A -31

B -32

C -33

D -34

Answer - D

Explanation

Here a = 7, d = 13 - 7 = 6, T_n = 205
Using formula T_n = a + (n - 1)d
T_n = 7 + (n - 1) x 6 = 205 where 205 is the n^th term.
=> 7 + 6n - 6 = 205
=> 6n = 205 - 1 
=> n = 204 / 6 
= 34

Q 4-^第6項が12で、^第8項が22の場合、APの最初の項は次のうちどれですか？

A --- 13

B -13

C -2

D -1

Answer - A

Explanation

Using formula T_n = a + (n - 1)d
T₆ = a + (6 - 1)d = 12   ...(i)
T₈ = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5
Using (i)
a = 12 - 5d 
= 12 - 25
= -13

Q 5-^第6項が12で、^第8項が22の場合、APの一般的な違いは次のうちどれですか？

A -4

B -5

C -6

D -7

Answer - B

Explanation

Using formula T_n = a + (n - 1)d
T₆ = a + (6 - 1)d = 12   ...(i)
T₈ = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5

Q 6-6^番目の項が12で、8^番目の項が22の場合、APの16^番目の項は次のうちどれですか？

A -60

B -61

C -62

D -63

Answer - C

Explanation

Using formula T_n = a + (n - 1)d
T₆ = a + (6 - 1)d = 12   ...(i)
T₈ = a + (8 - 1)d = 22   ...(ii)
Substract (i) from (ii)
=> 2d = 10 
=> d = 5
Using (i)
a = 12 - 5d 
= 12 - 25
= -13 
∴ T₁₆ = -13 + (16 - 1) x 5
= 75 - 13 
= 62

Q 7 -AP 5、9、13、17、...の最初の17項の合計は次のうちどれですか？

A -626

B -627

C -628

D -629

Answer - D

Explanation

Here a = 5, d = 9 - 5 = 4, n = 17
Using formula S_n = (n/2)[2a + (n - 1)d]
S₁₇ = (17/2)[2 x 5 + (17 - 1) x 4]
= (17/2)(10 + 64)
= 17 x 74 / 2
= 629

Q 8-シリーズ2、5、8、...、182の合計は次のうちどれですか？

A -5612

B -5613

C -5614

D -5615

Answer - A

Explanation

Here a = 2, d = 5 - 2 = 3, Tⁿ = 182
Using formula T_n = a + (n - 1)d
a + (n - 1)d = 182
=> 2 + (n - 1) x 3 = 182
=> 3n = 183
=> n = 61.
Using formula S_n = (n/2)[2a + (n - 1)d]
S₆₁ = (61/2)[2 x 2 + (61 - 1) x 3]
= (61/2)(4 + 180)
= 61 x 184 / 2
= 5612

Q 9-合計が15で、積が80の場合、APの3つの数値は何ですか？

A -5、7、3

B -2、5、8

C -6、7、2

D -5、5、5

Answer - B

Explanation

Let've numbers are a - d, a and a + d
Then a - d + a + a + d = 15
=> 3a = 15
=> a = 5
Now (a - d)a(a + d) = 80
=> (5 - d) x 5 x (5 + d) = 80
=> 25 - d² = 16
=> d² = 9
=> d = +3 or -3
∴ numbers are either 2, 5, 8 or 8, 5, 2.

Q 10 - GP 3、6、12、18の^第9^期は次のうちどれですか...？

A -766

B -768

C -772

D -774

Answer - B

Explanation

Here a = 3, r = 6 / 3 = 2, T₉ = ?
Using formula T_n = ar^{(n - 1)}
T₉ = 3 x 2^{(9 - 1)} 
=3 x 2⁸ 
=3 x 256
=768

Q 11-4^番目の項が54で、9^番目の項が13122の場合、GPの最初の項は次のうちどれですか？

A -2

B -3

C -4

D -6

Answer - A

Explanation

Using formula T_n = ar^{(n - 1)}
T₄ = ar^{(4 - 1)} = 54   
=> ar³ = 54   ...(i)
T₉ = ar^{(9 - 1)} = 13122
=> ar⁸ = 13122   ...(ii)
Dividing (ii) by (i)
=> r⁵ = 13122 / 54 = 243 = (3)⁵
=> r = 3
Using (i)
a x 27 = 54
=> a = 2

Q 12-4^番目の項が54で、9^番目の項が13122の場合、GPの一般的な比率は次のうちどれですか？

A -2

B -3

C -4

D -6

Answer - B

Explanation

Using formula T_n = ar^{(n - 1)}
T₄ = ar^{(4 - 1)} = 54   
=> ar³ = 54   ...(i)
T₉ = ar^{(9 - 1)} = 13122
=> ar⁸ = 13122   ...(ii)
Dividing (ii) by (i)
=> r⁵ = 13122 / 54 = 243 = (3)⁵
=> r = 3

Q 13-4^番目の項が54で、9^番目の項が13122の場合、GPの6^番目の項は^次のうちどれですか？

A -484

B -485

C -486

D -487

Answer - C

Explanation

Using formula T_n = ar^{(n - 1)}
T₄ = ar^{(4 - 1)} = 54   
=> ar³ = 54   ...(i)
T₉ = ar^{(9 - 1)} = 13122
=> ar⁸ = 13122   ...(ii)
Dividing (ii) by (i)
=> r⁵ = 13122 / 54 = 243 = (3)⁵
=> r = 3
Using (i)
a x 27 = 54
=> a = 2 
∴ T₆ = ar^{(6 - 1)} = 2 x (3)⁵  
= 2 x 243
= 486

Q 14-2つの数の合計は80です。最初の数の3倍が2番目の数の5倍と同じである場合、その数は何ですか？

A -50、30

B -60、20

C -70、10

D -65、15

Answer - A

Explanation

Let the numbers are y and 80 - y.
Then 3y = 5(80-y)
=> 8y = 400 
∴ y = 50
and second number = 80 - 50 = 30.

Q 15-3番目が5番目より16大きい場合、その数はいくつですか？

A -150

B -120

C -180

D -210

Answer - B

Explanation

Let the number be y.
Then (y / 3) - (y / 5) = 16
=> 5y - 3y = 16 x 15 = 240
=> 2y = 240
∴ y = 120

Q 16-合計が90の場合、3の3つの連続する倍数の中で最大の数はいくつですか？

A -21

B -30

C -33

D -36

Answer - C

Explanation

Let the numbers be 3y , 3y + 3, 3y + 6
Now 3y + 3y + 3 + 3y + 6 = 90
=> 9y = 81
=> y = 9
=> largest number = 3y + 6 = 3 x 9 + 6 
= 33

Q 17 -Findは、その15倍がその平方より16小さい場合、正の整数です。

A -13

B -14

C -15

D -16

Answer - D

Explanation

Let the positive integer by y.
Then y² - 15y = 16
=> y² - 15y - 16 = 0
=> y² - 16y + y - 16 = 0
=> y(y-16) + (y-16) = 0
=> (y+1)(y-16)= 0
∴ y = 16. as -1 is not a positive integer.

Q 18 -Findは、その23倍がその平方より63大きい場合、正の整数です。

A -7

B -8

C -9

D -10

Answer - A

Explanation

Let the positive integer by y.
Then 23y - 2y² = 63
=> 23y - 2y² - 63 = 0
=> 2y² - 23y + 63 = 0
=> 2y² - 14y - 9y + 63 = 0
=> 2y(y-7) - 9(y-7)= 0
=> (2y-9)(y-7)= 0
∴ y = 7. as 9/2 is not an integer.

Q 19-数値が3：2：5の比率であり、それらの2乗の合計が1862である場合、3つの数値のうち最小のものを見つけます。

A -13

B -14

C -12

D -11

Answer - B

Explanation

Let've number as 3y, 2y and 5y.
Then 9y² + 4y² + 25y² = 1862.
=> 38y² = 1862
=> y² = 1862 / 38 = 49
=> y = 7
∴ smallest number = 2y = 2 x 7 = 14.

Q 20-2桁の数字の合計は10です。数字を交換すると、取得した数字は元の数字より54少なくなります。何番ですか？

A -46

B -64

C -82

D -28

Answer - C

Explanation

Let the ten's digit is x and unit digit of number is y.
Then  x + y = 10   ...(i)
(10x + y) - (10y - x) = 54
=> 9x - 9y = 54
=> x - y = 6    ...(ii)
Adding (i) and (ii)
2x = 16
=> x = 8
Using (i)
y = 10 - x = 2
∴ number is 82.