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1.A can do 2/5 work in 10 days. And B completed the 1/3 work in 5 days. A worked for one day, then in how many days the remaining work done finished by A and B together.?
1.A can do 2/5 work in 10 days. And B completed the 1/3 work in 5 days. A worked for one day, then in how many days the remaining work done finished by A and B together.?
(A) 9 days (B)
Q.2) A certain number of men can complete a job in 30 days. If there were 5 men more, it could be completed in 10 days less. How many men were in the beginning?
рдХреБрдЫ рдирд┐рд╢реНрдЪрд┐рдд рдЖрджрдореА рдХрд┐рд╕реА рдХрд╛рдо рдХреЛ 30 рджрд┐рдиреЛрдВ рдореЗрдВ рдХрд░ рд╕рдХрддреЗ рд╣реИ. рдпрджрд┐ 5 рдЖрджрдореА рдФрд░ рд╣реЛрддреЗ рддреЛ рдХрд╛рдо 10 рджрд┐рди рдкрд╣рд▓реЗ реЩрддрдо рд╣реЛ рдЬрд╛рддрд╛. рддреЛ рд╢реБрд░реВ рдореЗрдВ рдХрд┐рддрдиреЗ рдЖрджрдореА рдереЗ?
(A) 10 (B) 15 (C) 20 (D) 25
Q.3) In a red fort, there was a sufficient food for 50 soldiers. If after 10 days 500 more CISF soldiers join the fort then the remaining food last for 35 days. So how many soldiers were in the beginning?
рд▓рд╛рд▓ рдХрд┐рд▓реЗ рдореЗрдВ 50 рджрд┐рдиреЛрдВ рдХреЗ рд▓рд┐рдП рдЦрд╛рдирд╛ рдкреНрд░рдпрд╛рдкреНрдд рд╣реИ рдпрджрд┐ 10 рджрд┐рди рдмрд╛рдж 500 CISF рд╕рд┐рдкрд╛рд╣реА рдФрд░ рдЖ рдЧрдП рддреЛ рд╢реЗрд╖ рдЦрд╛рдирд╛ рдХреЗрд╡рд▓ 35 рджрд┐рдиреЛрдВ рдХреЗ рд▓рд┐рдП рдкреНрд░рдпрд╛рдкреНрдд рд╣реЛрддрд╛ рд╣реИ рдЖрд░рдореНрдн рдореЗрдВ рд▓рд╛рд▓ рдХрд┐рд▓реЗ рдореЗрдВ рдХрд┐рддрдиреЗ рд╕рд┐рдкрд╛рд╣реА рдереЗ?
(A) 2500 (B) 3500 (C) 4000 (D) 4500
Q.4) If
рдпрджрд┐
(A) 9 (B) abc (C) a+b+c (D) 3abc
Q.5) If
рдпрджрд┐
(A) 180 (B) 198 (C) 234 (D) 252
Q.6) In a cyclic quadrilateral ABCD тИаBCD=120┬░ and passes through the centre of the circle. Then тИаABD =?
рдПрдХ рдЪрдХреНрд░реАрдп рдЪрддреБрд░реНрднреБрдЬ ABCD рдореЗрдВ , тИаBCD = 120┬░ рд╣реИрдВ рдФрд░ рдпрд╣ рд╡реГрддреНрдд рдХреЗ рдХреЗрдВрджреНрд░ рд╕реЗ рдЧреБрдЬрд░рддрд╛ рд╣реИ рддреЛ тИаABD =?
(A) 30┬░ (B) 40┬░ (C) 50┬░ (D) 60┬░
Q.7) The midpoints of AB and AC of a triangle ABC are X and Y respectively. If BC+XY=12 units, then BC-XY is
рддреНрд░рд┐рднреБрдЬ ABC рдореЗрдВ AB рдФрд░ AC рдХреА рдордзреНрдп рдмрд┐рдВрджреБ рдХреНрд░рдорд╢рдГ X рдФрд░ Y рд╣реИрдВред рдпрджрд┐ BC + XY = 12 рдЗрдХрд╛рдИ рд╣реИрдВ, рддреЛ BC тАУXY рд╣реИ-
(A) 10 units (B) 8 units (C) 6 units (D) 4 units
Q.8.) In an isosceles ╬ФABC, AD is the median to the unequal side meeting BC at D. DP is the angle bisector of тИаADB and PQ is drawn parallel to BC meeting AC at Q. Then the measure of тИаPDQ is
рдПрдХ рд╕рдорджреНрд╡рд┐рдмрд╛рд╣реБ ╬Ф ABC рдореЗрдВ, AD рдЕрд╕рдорд╛рди рднреБрдЬрд╛ BC рдХреА рдордзреНрдпрд┐рдХрд╛ рд╣реИрдВ рдЬреЛ D рдкрд░ рдорд┐рд▓рддреА рд╣реИрдВ| DP тИаADB рдХрд╛ рдХреЛрдг рджреНрд╡рд┐рднрд╛рдЬрдХ рд╣реИ рдФрд░ PQ рдХреЛ BC рдХреЗ рд╕рдорд╛рдВрддрд░ рдЦреАрдВрдЪрд╛ рдЬрд╛рддрд╛ рд╣реИ рдЬреЛ AC рд╕реЗ Q рдкрд░ рдорд┐рд▓рддреА рд╣реИрдВред рддреЛ тИаPDQ рдХрд╛ рдорд╛рди рд╣реИ тАУ
(A) 130┬░ (B) 90┬░ (C) 180┬░ (D) 45┬░
Q.9)129 meter from the foot of a cliff on level of ground, the angle of elevation of the top of a cliff is 30┬░. The height of this cliff is
рдЬрдореАрди рдХреЗ рд╕реНрддрд░ рдкрд░ рдПрдХ рдЪрдЯреНрдЯрд╛рди рдХреЗ рдкрд╛рдж рд╕реЗ 129 рдореАрдЯрд░ рдХреА рджреВрд░реА рд╕реЗ рдЪрдЯреНрдЯрд╛рди рдХреЗ рд╢реАрд░реНрд╖ рдХрд╛ рдЙрдирдпрди рдХреЛрдг 30 рдбрд┐рдЧреНрд░реА рд╣реИ | рдЗрд╕ рдЪрдЯреНрдЯрд╛рди рдХреА рдКрдВрдЪрд╛рдИ рд╣реИ-
(A) 50тИЪ3 metre (B) 45тИЪ3 metre (C) 43тИЪ3 metre (D) 47тИЪ3 metre
Q.10)The volume of metallic cylindrical pipe of uniform thickness is 748 c.c. Its length is 14 cm and its external radius is 9 cm. The thickness of the pipe is
рдПрдХ рд╕рдорд╛рди рдореЛрдЯрд╛рдИ рд╡рд╛рд▓реЗ рдзрд╛рддреБ рдХреЗ рдмреЗрд▓рдирд╛рдХрд╛рд░ рдкрд╛рдЗрдк рдХрд╛ рдЖрдпрддрди 748 рдШрди. рд╕реЗрдореА рд╣реИред рдЗрд╕рдХреА рд▓рдВрдмрд╛рдИ 14 рд╕реЗрдореА рд╣реИ рдФрд░ рдЗрд╕рдХреЗ рдмрд╛рд╣рд░реА рддреНрд░рд┐рдЬреНрдпрд╛ 9 рд╕реЗрдореА рд╣реИ рдкрд╛рдЗрдк рдХреА рдореЛрдЯрд╛рдИ рд╣реИ-
(A) 0.5 cm (B) 1.5 cm (C) 1 cm (D) 2 cm
Solution
Q.1)(A)
A can do 2/5 work in 10 days.
So A can do the whole work in 25 days.
B can do 1/3 work in 5 days.
So B can do the whole work in 15 days.
If A work for one day i.e 3 work has completed. So,
Q.2) (A)
Let the number of men originally = x.
Then,
So the remaining work will be completed in 9 days.
Q.3).(B)
Let the number of soldiers initially be x.
Then, after 10 days, the remaining food be = 40x.
Q.4) (D)
Q.5) (B)
Q.6) (C)
30┬░
Q.7) (D)
BC = 2XY
3XY = 12
XY= 4
BC = 8
BC тАУXY=4
Q.8) (B)
180/2 =90
Q.9) (B)
Q.10) (A)