# GRE Quantitative Practice Test With Solutions

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we give you here GRE Quantitative Practice Test With Solutions Top 10 Quantitative tips and tricks, and these numbers will normalize to our level of 5 with passing time. So, Enjoy!

### NUMBER CUBE

CUBES OF NUMBERS (TO MEMORISE):

SQUARES OF NUMBERS (TO MEMORISE):
1. CUBE ROOTS LAST DIGIT RULE: FROM THE TABLE IN POINT 1, NOTE THAT ALL CUBES END WITH THE SAME DIGIT AT THE UNIT’S PLACE, AS THEIR CUBE-ROOTS, EXCEPT FOR PAIRS OF 3 & 7 AND 2 & 8, WHICH END WITH EACH OTHER. THUS A NUMBER ENDING 2 HAS IT’S CUBE ENDING 8 AND VICE VERSA, AND A NUMBER ENDING 3 HAS IT’S CUBE ENDING 7 AND VICE VERSA.
2. SQUARE ROOT LAST DIGIT RULE:
3. PERFECT SQUARE LAST DIGIT RULE:FROM THE ABOVE TABLE WE CAN SEE THAT A PERFECT SQUARE CAN NEVER END IN A 2, 3, 7, OR 8
4.SOLVING CUBE ROOTS OF PERFECT CUBES:
WE TAKE A FEW KNOWN PERFECT CUBES. LET US FIRST TAKE 205379.
THE FIRST THING WE DO IS PUT A SLASH BEFORE THE LAST 3 DIGITS: 205 / 379.
FROM TABLE 1, THE CUBE ENDS IN 9, SO THE CUBE ROOT HAS TO END IN A 9, AND HENCE THE NUMBER IS _9.
THE PART TO THE LEFT OD THE SLASH IS 205, WHICH LIES BETWEEN 125 AND 216, THE CUBES OF 5 AND 6 RESPECTIVELY, AS CAN BE SEEN FROM TABLE 1. 5 IS THE SMALLER NUMBER OUT OF 5 AND 6, AND HENCE WE TAKE 5 AS THE TEN’S DIGIT OF THE CUBE ROOT. HENCE THE COMPLETE CUBE ROOT IS 59.
LET US NOW CONSIDER 300763. ADD A SLASH: 300 / 763. REFERRING TO TABLE 1, THE CUBE ROOT HAS TO END IN A 7, AND HENCE HAS TO HAVE THE FORM _7. THE NUMBER TO THE LEFT OF THE SLASH IS 300, WHICH IS BETWEEN 216 AND 343, THE CUBES OF 6 AND 7 RESPECTIVELY. SINCE, OF 6 AND 7, 6 IS THE SMALLER NUMBER, HENCE WE TAKE 6 AS THE TEN’S DIGIT OF THE CUBE ROOT. THUS THE COMPLETE CUBE ROOT IS 67.
FOR A CHANGE, LET US CONSIDER A 7 DIGIT NUMBER 1157625. ADD A SLASH: 1157 / 625.
REFERRING TO TABLE 1, THE CUBE ROOT HAS TO HAVE THE FORM _5.
THE REMAINING NUMBER 1157 IS BETWEEN 1000 AND 1331, THE CUBES OF 10 & 11 RESPECTIVELY. ACCEPTING THE SMALLER NUMBER 10, WE GET THE COMPLETE CUBE ROOT AS 105.
5. SOLVING SQUARE ROOTS OF PERFECT SQUARES:
LET US CONSIDER 7744.
SINCE THE SQUARE ENDS IN A 4, THE SQUARE ROOT HAS TO END IN A 2 OR AN 8, AS PER TABLE IN POINT 4.
NEXT, WE FIND THAT 7744 LIES BETWEEN TWO PERFECT SQUARES 6400 AND 8100, THE SQUARES OF 80 AND 90 RESPECTIVELY. HENCE THE SQUARE ROOT LIES BETWEEN 80 AND 90, OR IN OTHER WORDS IS EITHER 82 OR 88.
NEXT, WE NOTE THAT 7744 IS CLOSER TO THE BIGGER NUMBER 8100 THAN TO 6400, AND HENCE WE UNDERSTAND THAT THE ACTUAL SQUARE ROOT OF THE GIVEN NUMBER IS NOT 82, BUT 88.
LET US NOW CONSIDER 12544.
THE SQUARE ENDS IN A 4, AND HENCE THE SQUARE HAS TO BE OF THE FORM _2 OR _8.
NEXT, WE SEE THAT 12544 LIES BETWEEN 12100 AND 14400, THE SQUARES OF 110 AND 120 RESPECTIVELY. BUT 12544 IS CLOSER TO THE SMALLER NUMBER 12100 THAN TO 14400, AND HENCE THE SQUARE ROOT OF THE GIVEN NUMBER IS NOT 118, BUT 112.
6. MULTIPLICATION BY 99
FIRST ADD 2 ZEROS TO THE RIGHT OF THE NUMBER. THEN SUBTRACT THE ORIGINAL NUMBER FROM THIS NEW NUMBER, TO GET THE ANSWER.
EXAMPLE:
121*99 = 12100 – 121 = 11979
3857*99 = 385700 – 3857 = 381843
65*99 = 6500 – 65 = 6435
7. MULTIPLICATION BY 999
FIRST ADD 3 ZEROS TO THE RIGHT OF THE NUMBER. THEN SUBTRACT THE ORIGINAL NUMBER FROM THIS NEW NUMBER, TO GET THE ANSWER.
EXAMPLE:
121*999 = 121000 – 121 = 120879
3857*999 = 3857000 – 3857 = 3853143
65*999 = 65000 – 65 = 64935
8. MULTIPLICATION BY 9
FIRST ADD A ZERO TO THE RIGHT OF THE NUMBER. THEN SUBTRACT THE ORIGINAL NUMBER FROM THIS NEW NUMBER, TO GET THE ANSWER.
EXAMPLE:
121*9 = 1210 – 121 = 1089
3857*9 = 38570 – 3857 = 34713
65*9 = 650 – 65 = 585

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